A proton and a deutron ( q=+e, m=2.0 u) having same kinetic energies enter a region of uniform magnetic field B , moving perpendicular to B . The ratio of the radius r_d of deutron path to the radius r_p of the proton

Correct answer: A. 2 : 1 In uniform magnetic field, R= m qB = 2 m( K . E) qB Since same K.E aligned & R m q \& R_ deutron R_ proton = m_ d m_ p × q_ p q_ d \& = 2 × 1 \& γ_ d : γ_ p = 2 : 1 aligned

JEE Main 20249 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Evening Shift
Subject: Physics
Chapter: Magnetic Effects of Current

JEE Main 2024 · 9 April · Evening Shift · Physics · Magnetic Effects of Current

A proton and a deutron ( q=+e, m=2.0 u) having same kinetic...

A proton and a deutron (

q=+\mathrm{e}, m=2.0 \mathrm{u})

having same kinetic energies enter a region of uniform magnetic field

\vec{B}

, moving perpendicular to

\vec{B}

. The ratio of the radius

r_d

of deutron path to the radius

r_p

of the proton path is:

Options

A $\sqrt{2}: 1$
B $1: 1$
C $1: \sqrt{2}$
D $1: 2$

Correct answer

A. $\sqrt{2}: 1$

Step-by-step solution

In uniform magnetic field,

\mathrm{R}=\frac{\mathrm{m} \nu}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{ m}(\mathrm{ K} . \mathrm{E})}}{\mathrm{qB}}

Since same K.E

\begin{aligned}& \mathrm{R} \propto \frac{\sqrt{\mathrm{m}}}{\mathrm{q}} \& \therefore \frac{\mathrm{R}_{\text{deutron }}}{\mathrm{R}_{\text{proton }}}=\sqrt{\frac{\mathrm{m}_{\mathrm{d}}}{\mathrm{m}_{\mathrm{p}}}} \times \frac{\mathrm{q}_{\mathrm{p}}}{\mathrm{q}_{\mathrm{d}}} \& =\sqrt{2} \times 1 \& \therefore \gamma_{\mathrm{d}}: \gamma_{\mathrm{p}}=\sqrt{2}: 1\end{aligned}

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